Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int LG = 20;
vector<int> adj[100009 + 2], sparse[100009 + 2][LG + 2];
int n, m, q, lvl[100009 + 2];
int parent[100009 + 2][LG + 2];
vector<int> ad(vector<int> a, vector<int> b, int msz = 10) {
  vector<int> bye;
  int asz = a.size();
  int bsz = b.size();
  int sum = 0, i = 0, j = 0;
  while (i < asz && j < bsz && sum < msz) {
    if (a[i] <= b[j]) {
      bye.push_back(a[i]);
      i++;
      sum++;
    } else {
      bye.push_back(b[j]);
      j++;
      sum++;
    }
  }
  while (i < asz && sum < msz) {
    bye.push_back(a[i]);
    i++;
    sum++;
  }
  while (j < bsz && sum < msz) {
    bye.push_back(b[j]);
    j++;
    sum++;
  }
  return bye;
}
void dfs(int u, int par, int depth) {
  parent[u][0] = par;
  lvl[u] = depth;
  for (auto v : adj[u]) {
    if (v == par) continue;
    dfs(v, u, depth + 1);
  }
}
vector<int> lca(int u, int v, int koyta) {
  vector<int> bye;
  for (int i = LG; i >= 0; i--) {
    if (lvl[parent[u][i]] >= lvl[v]) {
      bye = ad(bye, sparse[u][i], koyta);
      u = parent[u][i];
    }
  }
  for (int i = LG; i >= 0; i--) {
    if (lvl[parent[v][i]] >= lvl[u]) {
      bye = ad(bye, sparse[v][i], koyta);
      v = parent[v][i];
    }
  }
  if (u == v) {
    bye = ad(bye, sparse[u][0], koyta);
    return bye;
  }
  for (int i = LG; i >= 0; i--) {
    if (parent[u][i] != parent[v][i]) {
      bye = ad(bye, sparse[u][i], koyta);
      bye = ad(bye, sparse[v][i], koyta);
      u = parent[u][i];
      v = parent[v][i];
    }
  }
  bye = ad(bye, sparse[u][0], koyta);
  bye = ad(bye, sparse[v][0], koyta);
  bye = ad(bye, sparse[parent[u][0]][0], koyta);
  return bye;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(parent, -1, sizeof parent);
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0, 1);
  for (int i = 1; i <= m; i++) {
    int city;
    cin >> city;
    sparse[city][0].push_back(i);
  }
  for (int j = 1; j <= LG; j++) {
    for (int i = 1; i <= n; i++) {
      if (parent[i][j - 1] != (-1)) {
        parent[i][j] = parent[parent[i][j - 1]][j - 1];
        sparse[i][j] = ad(sparse[i][j - 1], sparse[parent[i][j - 1]][j - 1]);
      }
    }
  }
  while (q--) {
    int u, v, koyta;
    cin >> u >> v >> koyta;
    vector<int> ses = lca(u, v, koyta);
    int sz = ses.size();
    cout << sz;
    for (int i = 0; i < sz; i++) {
      int x = ses[i];
      cout << " " << x;
    }
    cout << "\n";
  }
  return 0;
}